Output a matrix size n x m, 1 when the sum of the indices is even, 0 otherwise - matlab

Question

I'm attempting the following as a hobby, not as homework. In Computer Programming with MATLAB: J. Michael Fitpatrick and Akos Ledeczi, there is a practice problem that asks this:
Write a function called alternate that takes two positive integers, n and m, as input arguments (the function does not have to check the format of the input) and returns one matrix as an output argument. Each element of the n-by-m output matrix for which the sum of its indices is even is 1.
All other elements are zero.
A previous problem was similar, and I wrote a very simple function that does what it asks:
function A = alternate(n,m)
A(1:n,1:m)=0;
A(2:2:n,2:2:m)=1;
A(1:2:n,1:2:m)=1;
end
Now my question is, is that good enough? It outputs exactly what it asks for, but it's not checking for the sum. So far we haven't discussed nested if statements or anything of that sort, we just started going over very basic functions. I feel like giving it more functionality would allow it to be recycled better for future use.

Answer 1

Great to see you're learning, step 1 in learning any programming language should be to ensure you always add relevant comments! This helps you, and anyone reading your code. So the first improvement would be this:
function A = alternate(n,m)
% Function to return an n*m matrix, which is 1 when the sum of the indices is even
A(1:n,1:m)=0; % Create the n*m array of zeros
A(2:2:n,2:2:m)=1; % All elements with even row and col indices: even+even=even
A(1:2:n,1:2:m)=1; % All elements with odd row and col indicies: odd+odd=even
end
You can, however, make this more concise (discounting comments), and perhaps more clearly relate to the brief:
function A = alternate(n,m)
% Function to return an n*m matrix, which is 1 when the sum of the indices is even
% Sum of row and col indices. Uses implicit expansion (R2016b+) to form
% a matrix from a row and column array
idx = (1:n).' + (1:m);
% We want 1 when x is even, 0 when odd. mod(x,2) is the opposite, so 1-mod(x,2) works:
A = 1 - mod( idx, 2 );
end
Both functions do the same thing, and it's personal preference (and performance related for large problems) which you should use.
I'd argue that, even without comments, the alternative I've written more clearly does what it says on the tin. You don't have to know the brief to understand you're looking for the even index sums, since I've done the sum and tested if even. Your code requires interpretation.
It can also be written as a one-liner, whereas the indexing approach can't be (as you've done it).
A = 1 - mod( (1:n).' + (1:m), 2 ); % 1 when row + column index is even

Answer 2

Your function works fine and output the desired result, let me propose you an alternative:
function A = alternate(n,m)
A = zeros( n , m ) ; % pre-allocate result (all elements at 0)
[x,y] = meshgrid(1:m,1:n) ; % define a grid of indices
A(mod(x+y,2)==0) = 1 ; % modify elements of "A" whose indices verify the condition
end
Which returns:
>> alternate(4,5)
ans =
1 0 1 0 1
0 1 0 1 0
1 0 1 0 1
0 1 0 1 0
initialisation:
The first line is the equivalent to your first line, but it is the cannonical MATLAB way of creating a new matrix.
It uses the function zeros(n,m).
Note that MATLAB has similar functions to create and preallocate matrices for different types, for examples:
ones(n,m) Create
a matrix of double, size [n,m] with all elements set to 1
nan(n,m) Create a
matrix of double, size [n,m] with all elements set to NaN
false(n,m) Create a
matrix of boolean size [n,m] with all elements set to false
There are several other matrix construction predefined function, some more specialised (like eye), so before trying hard to generate your initial matrix, you can look in the documentation if a specialised function exist for your case.
indices
The second line generate 2 matrices x and y which will be the indices of A. It uses the function meshgrid. For example in the case shown above, x and y look like:
| x = | y = |
| 1 2 3 4 5 | 1 1 1 1 1 |
| 1 2 3 4 5 | 2 2 2 2 2 |
| 1 2 3 4 5 | 3 3 3 3 3 |
| 1 2 3 4 5 | 4 4 4 4 4 |
odd/even indices
To calculate the sum of the indices, it is now trivial in MATLAB, as easy as:
>> x+y
ans =
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
5 6 7 8 9
Now we just need to know which ones are even. For this we'll use the modulo operator (mod) on this summed matrix:
>> mod(x+y,2)==0
ans =
1 0 1 0 1
0 1 0 1 0
1 0 1 0 1
0 1 0 1 0
This result logical matrix is the same size as A and contain 1 where the sum of the indices is even, and 0 otherwise. We can use this logical matrix to modify only the elements of A which satisfied the condition:
>> A(mod(x+y,2)==0) = 1
A =
1 0 1 0 1
0 1 0 1 0
1 0 1 0 1
0 1 0 1 0
Note that in this case the logical matrix found in the previous step would have been ok since the value to assign to the special indices is 1, which is the same as the numeric representation of true for MATLAB. In case you wanted to assign a different value, but the same indices condition, simply replace the last assignment:
A(mod(x+y,2)==0) = your_target_value ;

Answer 3

I don't like spoiling the learning. So let me just give you some hints.
Matlab is very efficient if you do operations on vectors, not on individual elements. So, why not creating two matrices (e.g. N, M) that holds all the indices? Have a look at the meshgrid() function.
Than you might be able find all positions with an even sum of indices in one line.
Second hint is that the outputs of a logic operation, e.g. B = A==4, yields a logic matrix. You can convert this to a matrix of zeros by using B = double(B).
Have fun!